Question

Could i have some super quick quick help? graph part isn't necessary

Answer 1

y=4x²-1

#1

• y=(2x)²-1²
• y=(2x+1)(2x-1)

#2

Zeros are 1/2,-1/2

#3

Convert to vertex form y=a(x-h)²+k

• y=(4x+0)²-1

Vertex(h,k)=(0,-1)

#4

Graph attached

Answer 2

i)
`y=(2x+1)(2x-1)`

ii)
`x=-\frac12, x=\frac12`

iii) (0, -1)

iv) see attached

Explanation:

Given quadratic:
`y=4x^2-1`

Factored form

We can use the Difference of Two Squares to factor the given quadratic.

`a^2-b^2=(a+b)(a-b)`

Therefore,

`a^2=4x^2=(2x)^2\implies a=2x`

`b^2=1=1^2 \implies b=1`

So the relation in factored form is:

`y=(2x+1)(2x-1)`

Zeros

The zeros of the quadratic polynomial are the x-coordinates of the points where the graph intersects the x-axis, i.e. when y = 0

`\implies y=0`

`\implies (2x+1)(2x-1)=0`

`\implies (2x+1)=0\implies x=-\frac12`

`\implies (2x-1)=0\implies x=\frac12`

Therefore, the zeros are -1/2 and 1/2

Vertex

The x-coordinate of the vertex is the midpoint of the zeros.

`\textsf{midpoint}=(-\frac12+\frac12)/(2)=0`

To find the y-coordinate of the vertex, substitute the found value of x into the given equation:

`\implies y=4(0)^2-1=-1`

Therefore, the vertex is (0, -1)